JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 3)
Let $$x=2$$ be a local minima of the function $$f(x)=2x^4-18x^2+8x+12,x\in(-4,4)$$. If M is local maximum value of the function $$f$$ in ($$-4,4)$$, then M =
$$18\sqrt6-\frac{33}{2}$$
$$18\sqrt6-\frac{31}{2}$$
$$12\sqrt6-\frac{33}{2}$$
$$12\sqrt6-\frac{31}{2}$$
Explanation
$$
\begin{aligned}
& f(x)=8 x^3-36 x+8 \\\\
& =4\left(2 x^3-9 x+2\right) \\\\
& =4(x-2)\left(2 x^2+4 x-1\right) \\\\
& =4(x-2)\left(x-\frac{-2+\sqrt{6}}{2}\right)\left(x-\frac{-2 \sqrt{6}}{2}\right)
\end{aligned}
$$
Local maxima occurs at $x=\frac{-2+\sqrt{6}}{2}=x_0$
$$ f\left(x_0\right)=12 \sqrt{6}-\frac{33}{2} $$
Local maxima occurs at $x=\frac{-2+\sqrt{6}}{2}=x_0$
$$ f\left(x_0\right)=12 \sqrt{6}-\frac{33}{2} $$
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